- #1
ken.drea
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Question: A kicker on a football team attempts to kick a field goal from 35 yards. The attempt just clears the goal post that is 10 feet high. The football lands 8 feet behind the plane of the goal. What speed and angle was the football initially kicked? (Use acceleration of gravity as 32 ft/s^2).
Variables given&needed
a= 32ft/s^2
Vi = ?
X= 113 ft (35yrds = 105 ft, then +8ft)
Vyf= 0
At first, I tried plugging them into this equation: Vyf^2 = Vyi^2 + 2ay
I used 10 feet as the y value. I got 25.298 as my Vyi. Then, I solved for the angle by using sine. I got an error. Then I realized that 10 feet is not the full y value, It's only the height of the goal thing.
Now I'm stuck. I was going to find time, but I'm missing a lot of variables. I was going to use it into the formula "x=Vt" to find the velocity.
Note:
a = acceleration
Vi = velocity initial
Vf = velocity final
X = horizontal distance
Y = vertical distance
t =time
Variables given&needed
a= 32ft/s^2
Vi = ?
X= 113 ft (35yrds = 105 ft, then +8ft)
Vyf= 0
At first, I tried plugging them into this equation: Vyf^2 = Vyi^2 + 2ay
I used 10 feet as the y value. I got 25.298 as my Vyi. Then, I solved for the angle by using sine. I got an error. Then I realized that 10 feet is not the full y value, It's only the height of the goal thing.
Now I'm stuck. I was going to find time, but I'm missing a lot of variables. I was going to use it into the formula "x=Vt" to find the velocity.
Note:
a = acceleration
Vi = velocity initial
Vf = velocity final
X = horizontal distance
Y = vertical distance
t =time